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A router is emitting out IP packets whose total length (data plus header) is 1024 bytes.Assuming that packets live for 10 sec, what is the maximum line speed the router can operate at without danger of cycling through the IP datagram identification number space?

1. 53.6 Mbps
2. 3.5 Tbps
3. 5.36 Mbps
4. 35 Tbps
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there are 16 bits in the identification number space of an IPV4 datagram

since it is said the lifetime of a packet is 10 sec. So in order to prevent wrap around, at max we can send 216  packets in 10seconds, i.e. in 1 sec, we can send 216/10 packets.

length of a packet is given as 1024B. So in 1 sec, we can send 216 * 1024/10 Bytes. So in 1 sec, we can send 6.4MB. So maximum data rate that is possible is 6.4MBps= (6.4*8)Mbps= 51.2Mbps.

What is the answer for this?

Total length of 1 packet=1024 bytes.

Packet lifetime is 10 seconds , means if a packet with Identification number 1 is produced at time "t" then only at t+10 the same packet should be produced.

Idenfication number space available is $2^{16}$

Means, the router should be able to exhaust all Identification numbers in 10 seconds only.

1 Identification number is given to a packet of size->1024 Bytes.

So, Router transmission rate must be $\frac{2^{16} \times 1024 \times 8}{10}=53.68Mbps$
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